How does the structure of the alveoli relate to its function in the lungs

Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq

andrew-mc [135]

The force on each balloon is 2×10^−3 N.

Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

$\sum F_{y} = Tcos10\textdegree - mg = 0\\\\T = \frac{mg}{cos10\textdegree } \\\\\sum F_{y} = Tsin10\textdegree - mg = 0\\$

So,

$F_{e} = \frac{mg}{cos10\textdegree }sin10\textdegree = mgtan10\textdegree \\\\= (0.00100kg)(9.8m/s^{2})tan10\textdegree \\\\F_{e} = 2 \times 10^{-3}N$

A force is an influence that can change the motion of an object. A force can cause an object with mass to change its velocity (e.g. moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newton (N).

Learn more about force here:

brainly.com/question/13191643

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3 0

1 year ago

A total distance of 4.5km. The overall journey takes 0.62h, whats the average velocity

Westkost [7]

Answer:

7.3km/hr

Explanation:

v=d/t=4.5km/0.62hrs=7.3km/hr

6 0

3 years ago

You plug your microwave into an outlet and then you heat up a piece of pizza in it. This is an example of an...

dem82 [27]

Answer:

b

Explanation:

i think we have not learned that yet

im so sorry if it is wrong

3 0

2 years ago

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

$P=\frac{\frac{1}{2}mv^2}{t}=const.$

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

$\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}$

where:

$t_1=1.40 s$ is the time the car needs to accelerates to $v_1=28.0 mph$

$t_2$ is the time the car needs to accelerate to $v_2=57.0 mph$

Therefore, solving for $t_2$ ,

$t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s$

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity

$v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s$ is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

$a=\frac{12.5-0}{1.40}=8.9 m/s^2$

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

$F=ma$